Solve 1/2x-3+1/x-5=1

$$\frac { 1 } { 2 x - 3 } + \frac { 1 } { x - 5 } = 1$$

Solve for x

$$x=\frac{3\sqrt{2}}{2}+4\approx 6.121320344$$

$$x=-\frac{3\sqrt{2}}{2}+4\approx 1.878679656$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values \frac{3}{2},5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(2x-3\right), the least common multiple of 2x-3,x-5.

$$x-5+2x-3=\left(x-5\right)\left(2x-3\right)$$

Combine x and 2x to get 3x.

$$3x-5-3=\left(x-5\right)\left(2x-3\right)$$

Subtract 3 from -5 to get -8.

$$3x-8=\left(x-5\right)\left(2x-3\right)$$

Use the distributive property to multiply x-5 by 2x-3 and combine like terms.

$$3x-8=2x^{2}-13x+15$$

Subtract 2x^{2} from both sides.

$$3x-8-2x^{2}=-13x+15$$

Add 13x to both sides.

$$3x-8-2x^{2}+13x=15$$

Combine 3x and 13x to get 16x.

$$16x-8-2x^{2}=15$$

Subtract 15 from both sides.

$$16x-8-2x^{2}-15=0$$

Subtract 15 from -8 to get -23.

$$16x-23-2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+16x-23=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 16 for b, and -23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-2\right)\left(-23\right)}}{2\left(-2\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-2\right)\left(-23\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-16±\sqrt{256+8\left(-23\right)}}{2\left(-2\right)}$$

Multiply 8 times -23.

$$x=\frac{-16±\sqrt{256-184}}{2\left(-2\right)}$$

Add 256 to -184.

$$x=\frac{-16±\sqrt{72}}{2\left(-2\right)}$$

Take the square root of 72.

$$x=\frac{-16±6\sqrt{2}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-16±6\sqrt{2}}{-4}$$

Now solve the equation x=\frac{-16±6\sqrt{2}}{-4} when ± is plus. Add -16 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}-16}{-4}$$

Divide -16+6\sqrt{2} by -4.

$$x=-\frac{3\sqrt{2}}{2}+4$$

Now solve the equation x=\frac{-16±6\sqrt{2}}{-4} when ± is minus. Subtract 6\sqrt{2} from -16.

$$x=\frac{-6\sqrt{2}-16}{-4}$$

Divide -16-6\sqrt{2} by -4.

$$x=\frac{3\sqrt{2}}{2}+4$$

The equation is now solved.

$$x=-\frac{3\sqrt{2}}{2}+4 x=\frac{3\sqrt{2}}{2}+4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values \frac{3}{2},5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(2x-3\right), the least common multiple of 2x-3,x-5.

$$x-5+2x-3=\left(x-5\right)\left(2x-3\right)$$

Combine x and 2x to get 3x.

$$3x-5-3=\left(x-5\right)\left(2x-3\right)$$

Subtract 3 from -5 to get -8.

$$3x-8=\left(x-5\right)\left(2x-3\right)$$

Use the distributive property to multiply x-5 by 2x-3 and combine like terms.

$$3x-8=2x^{2}-13x+15$$

Subtract 2x^{2} from both sides.

$$3x-8-2x^{2}=-13x+15$$

Add 13x to both sides.

$$3x-8-2x^{2}+13x=15$$

Combine 3x and 13x to get 16x.

$$16x-8-2x^{2}=15$$

Add 8 to both sides.

$$16x-2x^{2}=15+8$$

Add 15 and 8 to get 23.

$$16x-2x^{2}=23$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+16x=23$$

Divide both sides by -2.

$$\frac{-2x^{2}+16x}{-2}=\frac{23}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{16}{-2}x=\frac{23}{-2}$$

Divide 16 by -2.

$$x^{2}-8x=\frac{23}{-2}$$

Divide 23 by -2.

$$x^{2}-8x=-\frac{23}{2}$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-\frac{23}{2}+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-\frac{23}{2}+16$$

Add -\frac{23}{2} to 16.

$$x^{2}-8x+16=\frac{9}{2}$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=\frac{9}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{\frac{9}{2}}$$

Simplify.

$$x-4=\frac{3\sqrt{2}}{2} x-4=-\frac{3\sqrt{2}}{2}$$

Add 4 to both sides of the equation.

$$x=\frac{3\sqrt{2}}{2}+4 x=-\frac{3\sqrt{2}}{2}+4$$